NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Q: What is the significance of Argand Plane in Class 11 Maths Chapter 5?

The Argand Plane is significant in Class 11 Maths Chapter 5 (Complex Numbers) as it provides a geometric representation of complex numbers. Each complex number z=x+iy is plotted as a point (x,y) in a 2D plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. This visual approach helps understand complex number operations like addition, subtraction, modulus, and conjugation more intuitively.

Q: What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 5?

Important topics covered in NCERT Exemplar Class 11 Maths Chapter 5 (Complex Numbers and Quadratic Equations) include: Algebra of complex numbers (addition, subtraction, multiplication, division) Modulus and argument of complex numbers Polar and exponential form Geometrical representation on Argand Plane Solving quadratic equations with complex roots Use of identities and inequalities involving complex numbers

NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

Edited By Komal Miglani | Updated on Mar 31, 2025 01:36 AM IST

NCERT Exemplar Class 11 Maths Solutions chapter 5 covers Complex numbers and their applications. In our daily lives, we come across many concepts involving real and imaginary components, such as electrical circuits and signal processing. So, what is a complex number? A complex number is a number that has two parts: a real part and an imaginary part. It is written in the form a + bi, where a is the real part and b is the imaginary part with i² = -1. Complex numbers are useful in solving equations that have no real solutions. We have been studying complex numbers over the years, with any ‘real number’ being said to be a complex number, which can be denoted as a variable or alphabet in simpler terms. The other number, along with the real number, is a part of the complex number, and is called the ‘imaginary number.’ Both of them appear together in a problem.

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NCERT Solutions for Class 11 Mathematics Chapters
Importance of solving NCERT questions for class 11, Chapter 5, complex numbers and quadratic equations:
NCERT Exemplar Class 11 Solutions":

Terms such as integer, conjugate, square root, polar, etc. will be studied in NCERT Exemplar Class 11 Maths Solutions Chapter 5, which will be utilised to find results for these complex number problems. Unchanging practice NCERT Solutions for Class 11 via a worksheet and exercise is highly recommended for students preparing for a tough examination, as it contributes to a deep understanding of the subject and enables them to perform analogous tests.

Class 11 Maths chapter 5 solutions Exercise: 5.3

Page number: 91-97

Total questions: 50

Question:1 For a positive integer n, find the value of $(1 - i)^{n}$ $(1 - 1 / i)^{n}$

Answer:

$(1 - i)^{n} (1 - 1 / i)^{n}$
$= (1 - i)^{n} (1 + i)^{n}$
$= (1 - i^{2})^{n}$
$= 2^{n}$

Question:2 Evaluate $\sum_{i = 1}^{13}$ $(i^{n} + i^{n + 1})$ where n $ϵ$ N.

Answer:

$\sum_{i = 1}^{13}$ $(i^{n} + i^{n + 1})$
$=$ $\sum_{i = 1}^{13}$ $(1 + i) i^{n}$
$= (1 + i) (1 + i^{2} + i^{3} + i^{4} . . . . + i^{12} + i^{13})$
$= (1 + i) \frac{i (i^{13} - 1)}{(i - 1)}$
$= (1 + i) i \frac{(i - 1)}{(i - 1)}$
$= (1 + i) i$
$= i + i^{2}$
$= i - 1$

Question:3 If ${\frac{(1 + i)}{(1 - i)}}^{3} - {\frac{(1 - i)}{(1 + i)}}^{3} = x + i y$ then find (x, y).

Answer:

Given ${(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = x + i y$
$= {(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3}$
$= {(\frac{1 + 2 i + i^{2}}{1 - i^{2}})}^{3} - {(\frac{1 - 2 i + i^{2}}{1 - i^{2}})}^{3}$
$= (\frac{2 i}{2})^{3} - (\frac{- 2 i}{2})^{3}$
$= i^{3} - (- i^{3})$
$= 2 i^{3} = 0 - 2 i$
$T h u s, (x, y) = (0, - 2)$

Question:4 If $(1 + i)^{2} / (2 - i) = x + i y$ then find the value of x + y.

Answer:

$x + i y = \frac{(1 + i)^{2}}{(2 - i)} = \frac{(1 + 2 i + i^{2})}{(2 - i)} = \frac{2 i}{(2 - i)}$

$O n r a t i o n a l i s i n g 2 i (2 + i) / (2 - i) (2 + i)$ ,
$= \frac{(4 i + 2 i^{2})}{(4 - i^{2})} = (\frac{(4 i - 2)}{(4 + 1)})$
$= \frac{- 2}{5} + \frac{4 i}{5}$
$x = \frac{- 2}{5} a n d y = \frac{4}{5}$

$S o, x + y = - \frac{2}{5} + \frac{4}{5} = \frac{2}{5}$

Question:5 If ${(\frac{1 - i}{1 + i})}^{100} = a + i b$ then find (a, b).

Answer:

$a + i b = {(\frac{1 - i}{1 + i})}^{100} = {[\frac{1 - i}{1 + i} * \frac{1 - i}{1 - i}]}^{100}$
$= {[\frac{(1 - i)^{2}}{1 - i^{2}}]}^{100}$
$= {(\frac{(1 - 2 i + i^{2})}{(1 + 1)})}^{100}$
$= {(\frac{- 2 i}{2})}^{100}$
$= (i^{4})^{25} = 1$
$H e n c e, (a, b) = (1, 0)$

Question:6 If $a = \cos θ + i \sin θ$ , find the value of $\frac{(1 + a)}{(1 - a)}$ .

Answer:

$a = \cos θ + i \sin θ$
$\frac{(1 + a)}{(1 - a)}$ $= \frac{(1 + c o s θ) + i \sin θ}{(1 - c o s θ) - i s i n θ}$

$= \frac{(2 \cos^{2} \frac{θ}{2} + i 2 \sin \frac{θ}{2} \cos \frac{θ}{2})}{(2 \sin^{2} \frac{θ}{2} - i 2 \sin \frac{θ}{2} \cos \frac{θ}{2})}$
$= \frac{2 \cos \frac{θ}{2} (\cos \frac{θ}{2} + i \sin \frac{θ}{2})}{2 \sin \frac{θ}{2} (\sin \frac{θ}{2} - i \cos \frac{θ}{2})}$
$= \frac{i \cos \frac{θ}{2} (\cos \frac{θ}{2} + i \sin \frac{θ}{2})}{\sin \frac{θ}{2} (i \sin \frac{θ}{2} - i^{2} \cos \frac{θ}{2})}$
$= \frac{i \cos \frac{θ}{2} (\cos \frac{θ}{2} + i \sin \frac{θ}{2})}{\sin \frac{θ}{2} (i \sin \frac{θ}{2} + \cos \frac{θ}{2})}$
$= i \cot \frac{θ}{2}$

Question:7 If $(1 + i) z = (1 - i) \bar{z},$ then show that $z = - i \bar{z}$

Answer:

$(1 + i) z = (1 - i) \bar{z},$
$z = \frac{1 - i}{1 + i} \bar{z}$
Rationalising the denominator,
$\frac{(1 - i) (1 - i)}{(1 + i) (1 - i)} \bar{z}$
$= \frac{{(1 - i)}^{2}}{1 - i^{2}} \bar{z}$
$= \frac{(1 - 2 i + i^{2})}{1 + 1} \bar{z}$
$= \frac{(1 - 2 i - 1)}{2} \bar{z}$
$= - i \bar{z}$
Hence, Proved.

Question:8 If $z = x + i y$ , then show that $z \bar{z} + 2 (z + \bar{z}) + b = 0$ where b belongs to R, representing z in the complex plane as a circle.

Answer:

$z = x + i y$
$\bar{z} = x - i y$
$N o w, w e a l s o h a v e, z \bar{z} + 2 (z + \bar{z}) + b = 0$
$(x + i y) (x - i y) + 2 (x + i y + x - i y) + b = 0$
This equation is an equation of a circle.

Question:9 If the real part of $\frac{(\bar{z} + 2)}{(\bar{z} - 1)}$ is 4, then show that the locus of the point representing z in the complex plane is a circle.

Answer:

$L e t z = x + i y$
$\frac{\bar{z} + 2}{\bar{z} - 1} = \frac{x - i y + 2}{x - i y - 1}$
$= \frac{[(x + 2) - i y] [(x - 1) + i y]}{[(x - 1) - i y] [(x - 1) + i y]}$
$= \frac{(x - 1) (x + 2) + y^{2} + i [(x + 2) y - (x - 1) y]}{(x - 1)^{2} + y^{2}}$
$r e a l p a r t = 4 \Rightarrow \frac{(x - 1) (x + 2) + y^{2}}{(x - 1)^{2} + y^{2}} = 4$
$x^{2} + x - 2 + y^{2} = 4 (x^{2} - 2 x + 1 + y^{2})$
$3 x^{2} + 3 y^{2} - 9 x + 6 = 0$

The equation obtained is that of a circle. Hence, the locus of z is a circle.

Question:10 Show that the complex number z, satisfying the condition $a r g (\frac{z - 1}{z + 1}) = \frac{π}{4}$ lies on a circle.

Answer:

Let z=x+iy
$a r g (\frac{z - 1}{z + 1}) = \frac{π}{4}$
$a r g (z - 1) - a r g (z + 1) =$ $\frac{π}{4}$
$a r g (x + i y - 1) - a r g (x + i y + 1) =$ $\frac{π}{4}$
$a r g (x - 1 + i y) - a r g (x + 1 + i y) =$ $\frac{π}{4}$
$\tan^{- 1} \frac{y}{x - 1} - \tan^{- 1} \frac{y}{x + 1} =$ $\frac{π}{4}$
$\tan^{- 1} \frac{\frac{y}{x - 1} - \frac{y}{x + 1}}{1 + (\frac{y}{x - 1}) (\frac{y}{x + 1})} =$ $\frac{π}{4}$
$\tan^{- 1} (\frac{y (x + 1 + - x + 1)}{x^{2} - 1 + y^{2}}) = \frac{π}{4}$
$\tan \frac{π}{4} = \frac{2 y}{x^{2} + y^{2} - 1}$
$x^{2} + y^{2} - 1 = 2 y$
$x^{2} + y^{2} - 1 - 2 y = 0$
The equation obtained represents the equation of a circle

Question:11 Solve that equation $| z | = z + 1 + 2 i$ .

Answer:

$| z | = z + 1 + 2 i$
$S u b s t i t u t i n g z = x + i y w e g e t | x + i y | = x + i y + 1 + 2 i$
$| z | = \sqrt{(x^{2} + y^{2})} = (x + 1) + i (y + 2)$
Comparing real and imaginary parts $\sqrt{(x^{2} + y^{2})} = (x + 1)$
And $0 = y + 2, y = - 2$
Substituting the value of y in $\sqrt{(x^{2} + y^{2})} = (x + 1)$
$x^{2} + (- 2)^{2} = (x + 1)^{2}$
$x^{2} + 4 = x^{2} + 2 x + 1$

Hence, $x = \frac{3}{2}$
Hence, $z = x + i y$
$= \frac{3}{2} - 2 i$

Question:12 If $| z + 1 | = z + 2 (1 + i)$ , then find z

Answer:

We have $| z + 1 | = z + 2 (1 + i)$
Substituting $z = x + i y$ we get $x + i y + 1 = x + i y + 2 i + 1$
$| z | = \sqrt{(x^{2} + y^{2})} = \sqrt{(x + 1)^{2} + y^{2}} = (x + 2) + i (y + 2)$
Comparing real and imaginary parts, $\sqrt{(x + 1)^{2} + y^{2}} = (x + 2)$
$a n d 0 = y + 2; y = - 2$
Substituting the value of y in $\sqrt{(x + 1)^{2} + y^{2}} = (x + 2)$
$(x + 1)^{2} + (- 2)^{2} = (x + 2)^{2}$
$x^{2} + 2 x + 1 + 4 = x^{2} + 4 x + 4$
$2 x = 1$
Hence, $x = \frac{1}{2}$
Thus, $z = x + i y = \frac{1}{2} - 2 i$

Question:13 If $a r g (z - 1) = a r g (z + 3 i)$ , then find x – 1 : y. where $z = x + i y$

Answer:

Given that $a r g (z - 1) = a r g (z + 3 i)$
$a r g (x + i y - 1) = a r g (x + i y + 3 i)$
$a r g (x - 1 + i y) = a r g (x + i (y + 3))$
$\tan^{- 1} \frac{y}{x - 1} = \tan^{- 1} \frac{y + 3}{x}$
$\frac{y}{x - 1} = \frac{y + 3}{x}$
$x y = x y - y + 3 x - 3$
$3 x - 3 = y$
$\frac{x - 1}{y} = \frac{1}{3}$

Question:14 Show that $| \frac{z - 2}{z - 3} | = 2$ represents a circle. Find its centre and radius.

Answer:

$| \frac{z - 2}{z - 3} | = 2$ Substituting $z = x + i y$ , we get $| \frac{x + i y - 2}{x + i y - 3} | = 2$
$| x - 2 + i y | = 2 | x - 3 + i y |$
$\sqrt{(x - 2)^{2} + y^{2}} = 2 \sqrt{((x - 3)^{2} + y^{2}}$
$x^{2} - 4 x + 4 + y^{2} = 4 (x^{2} - 6 x + 9 + y^{2})$
$3 x^{2} + 3 y^{2} - 20 x + 32 = 0$
$x^{2} + y^{2} - \frac{20}{3} x + \frac{32}{} 3 = 0$
$(x - \frac{10}{3})^{2} + y^{2} + \frac{32}{3} - \frac{100}{9} = 0$
Thus, the centre of circle is $(\frac{10}{3}, 0)$ and radius is $2 / 3$

Question:15 If $\frac{z - 1}{z + 1}$ is a purely imaginary number $(z \neq - 1)$ , then find the value of $| z |$ .

Answer:

Let $z = x + i y$
$\frac{z - 1}{z + 1} = \frac{x + i y - 1}{x + i y + 1}$
$\frac{[(x - 1) + i y] [(x + 1) - i y]}{[(x + 1) + i y] [(x + 1) - i y]}$
$= \frac{(x - 1) (x + 1) + y^{2} + i [(x + 1) y - (x - 1) y]}{(x + 1)^{2} + y^{2}}$
$\frac{z - 1}{z - 1}$ is purely imaginary
$\frac{(x - 1) (x + 1) + y^{2}}{(x + 1)^{2} + y^{2}} = 0$
$x^{2} - 1 + y^{2} = 0$
$x^{2} + y^{2} = 1$ Hence, $| z | = 1$

Question:16 z₁ and z₂ are two complex numbers such that $| z_{1} | = | z_{2} |$ and $a r g (z_{1}) + a r g (z_{2}) = π$ , then show that $| z_{1} | = - | z_{2} |$ .

Answer:

Let $z_{1} = | z_{1} | (c o s θ_{1} + i s i n θ_{1})$ and $z_{2} = | z_{2} | (c o s θ_{2} + i s i n θ_{2})$
Given that $| z_{1} | = | z_{2} |$
And $a r g (z_{1}) + a r g (z_{2}) = π$

$θ_{1} + θ_{2} = π$

$θ_{1} = π - θ_{2}$

$z_{2} = | z_{2} | (c o s θ_{2} + i s i n θ_{2})$

$z_{1} = | z_{2} | (- \cos θ_{2} + i s i n θ_{2})$

$z_{1} = - | z_{2} | (\cos θ_{2} - i s i n θ_{2})$

$z_{1} = - | z_{2} | (\cos θ_{2} - i s i n θ_{2})$ , $| z_{1} | = - | z_{2} |$

Question:17 If |z₁| = 1 (z₁ ≠ –1) and $z_{2} = \frac{z_{1} - 1}{z_{1} + 1}$ then show that the real part of z₂ is zero

Answer:

Let $z_{1} = x + i y$ $| z_{1} | = \sqrt{x^{2} + y^{2}} = 1$
$z_{2} = \frac{z_{1} - 1}{z_{1} + 1} = \frac{x + i y - 1}{x + i y + 1}$
$= \frac{[(x - 1) + i y] [(x + 1) - i y]}{[(x + 1) + i y] [(x + 1) - i y]}$
$= \frac{(x - 1) (x + 1) + y^{2} + i [(x + 1) y - (x - 1) y]}{(x + 1)^{2} + y^{2}}$
$\frac{x^{2} + y^{2} - 1 + 2 i y}{(x + 1)^{2} + y^{2}} = 0$
Since, $x^{2} + y^{2} = 1$
$\frac{1 - 1 + 2 i y}{(x + 1)^{2} + y^{2}} = \frac{0 + 2 i y}{(x + 1)^{2} + y^{2}}$
Therefore, the real part of $z_{2}$ is zero

Question:18 If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, then find $a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}})$

Answer:

$z_{1}$ and $z_{2}$ are conjugate complex numbers.
The negative side of the real axis $= r_{1} (c o s θ - i s i n θ)$
$= r_{1} (\cos (- θ_{1}) + i \sin (- θ_{1}))$
Similarly, $z_{3} = r_{2} (\cos (θ_{2}) - i \sin (θ_{2}))$
$z_{4} = r_{2} (\cos (- θ_{2}) + i \sin (- θ_{2}))$
$a r g (\frac{z_{1}}{z_{4}}) + a r g (\frac{z_{2}}{z_{3}})$ $= a r g (z_{1}) - a r g (z_{4}) + a r g (z_{2}) - a r g (z_{3}) ? ?$
$= θ_{1} - (- θ_{2}) + (- θ_{1}) - θ_{2} = 0$

Question:19 If $| z_{1} | = | z_{2} | = . . . = | z_{n} | = 1$ , then
Show that $| z_{1} + z_{2} + z_{3} + . . . + z_{n} | = | 1 / z_{1} + 1 / z_{2} + . . . + 1 / z_{n} |$

Answer:

$| z_{1} | = | z_{2} | = . . . = | z_{n} | = 1$
$| z_{1} |^{2} = | z_{2} |^{2} = . . . = | z_{n} |^{2} = 1$
$z_{1} \bar{z_{1}} = z_{2} \bar{z_{2}} = . . . = z_{n} \bar{z_{n}}$
$z_{1} = \frac{1}{\bar{z_{1}}}, z_{2} = \frac{1}{\bar{z_{2}}}, . . . z_{n} = \frac{1}{\bar{z_{n}}}$
Now, $| z_{1} + z_{2} + z_{3} + . . . z_{n} |$ $= | \frac{z_{1} \bar{z_{1}}}{\bar{z_{1}}} + \frac{z_{2} \bar{z_{2}}}{\bar{z_{2}}} + . . . + \frac{z_{n} \bar{z_{n}}}{\bar{z_{n}}} |$
= $| \frac{1}{\bar{z_{1}}} + \frac{1}{\bar{z_{2}}} + . . . + \frac{1}{\bar{z_{n}}} |$
$= | \frac{1}{z_{1}} + \frac{1}{z_{2}} + . . . + \frac{1}{z_{n}} |$

Question:20 If for complex numbers z₁ and z₂, $a r g (z_{1}) - a r g (z_{2}) = 0$ , then show that $| z_{1} - z_{2} | = | z_{1} | - | z_{2} |$

Answer:

Let $z_{1} = | z_{1} | (\cos θ_{1} + i \sin θ_{1})$ and $z_{2} = | z_{2} | (\cos θ_{2} + i \sin θ_{2})$
$a r g (z_{1}) - a r g (z_{1}) = 0$
$θ_{1} - θ_{2} = 0$
$θ_{1} = θ_{2}$
$z_{1} - z_{2} = (z_{1} | c o s θ_{1} - | z_{2} | c o s θ_{1}) + i (z_{1} | s i n θ_{1} - | z_{2} | s i n θ_{1})$
$z_{1} - z_{2} = \sqrt{(z_{1} | c o s θ_{1} - | z_{2} | c o s θ_{1})^{2} + (z_{1} | s i n θ_{1} - | z_{2} | s i n θ_{1})^{2}}$
$= \sqrt{| z_{1} |^{2} + | z_{2} |^{2} - 2 | z_{1} | | z_{2} | \cos^{2} θ_{1} - 2 | z_{1} | | z_{2} | \sin^{2} θ_{1}}$
$= \sqrt{| z_{1} |^{2} + | z_{2} |^{2} - 2 | z_{1} | | z_{2} |}$
$= \sqrt{{(| z_{1} | - | z_{2} |)}^{2}} = | z_{1} | - | z_{2} |$

Question:21 Solve the system of equations $R e (z^{2}) = 0, | z | = 2.$

Answer:

$R e (z^{2}) = 0, | z | = 2.$
$L e t z = x + i y$
$| z | = \sqrt{x^{2} + y^{2}} = 2$
$x^{2} + y^{2} = 4$
$z^{2} = x^{2} + 2 i x y - y^{2}$
$= (x^{2} - y^{2}) + 2 i x y$
$N o w, R e (z^{2}) = 0$
$x^{2} - y^{2} = 0$
$x^{2} = y^{2} = 2$
$x = y = \pm \sqrt{2}$
Hence, $z = x + i y = \pm \sqrt{2} \pm i \sqrt{2}$
$= \sqrt{2} + i \sqrt{2}, \sqrt{2} - i \sqrt{2}, - \sqrt{2} + i \sqrt{2} a n d - \sqrt{2} - i \sqrt{2}$

Question:22 Find the complex number satisfying the equation $z + \sqrt{2} | (z + 1) | + i = 0$ .

Answer:

$z + \sqrt{2} | (z + 1) | + i = 0 . . . (i)$
Substituting $z = x + i y$ we get $x + i y + \sqrt{2} | x + i y + 1 | + i = 0$
$x + i (1 + y) + \sqrt{2} [\sqrt{(x + 1)^{2} + y^{2}}] = 0$
$x + i (1 + y) + \sqrt{2} [\sqrt{(x^{2} + 2 x + 1 + y^{2})}] = 0$
1+y = 0
y = -1
$x + \sqrt{2} \sqrt{x^{2} + 2 x + 2} = 0$
$\sqrt{2} \sqrt{x^{2} + 2 x + 2} = - x$
$2 x^{2} + 4 x + 4 = x^{2}$
$x^{2} + 4 x + 4 = 0$
$(x + 2)^{2} = 0$
$x = - 2$
Hence, $z = x + i y = - 2 - i$

Question:23 Write the complex number $z = \frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$ in polar form

Answer:

$z = \frac{1 - i}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$
$= \frac{\sqrt{2} [\frac{1}{\sqrt{2}} - i 1 / \sqrt{2}]}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$
$= \frac{\sqrt{2} [\cos \frac{π}{4} - i s i n \frac{π}{4}]}{\cos \frac{π}{3} + i \sin \frac{π}{3}}$
$= \sqrt{2} [\cos (- \frac{π}{4} - \frac{π}{3}) + i \sin (- \frac{π}{4} - \frac{π}{3})]$
$= \sqrt{2} [\cos (- \frac{7 π}{12}) + i \sin (- \frac{7 π}{12})]$
$= - \sqrt{2} [\cos (\frac{5 π}{12}) + i \sin (\frac{5 π}{12})]$

Question:24 If z and w are two complex numbers such that $| z w | = 1$ and $a r g (z) - a r g (w) = π / 2$ , then show that $z w = - i$ .

Answer:

Let $z = | z | (\cos θ_{1} + i s i n θ_{1})$ and $w = | w | (\cos θ_{2} + i s i n θ_{2})$
$| z w | = | z | | w | = 1$
Also $a r g (z) - a r g (w) = \frac{π}{2}$
$θ_{1} - θ_{2} = \frac{π}{2}$
$\bar{z} w = | z | (\cos θ_{1} - i \sin θ_{1}) w$
$w = | w | (\cos θ_{2} + i \sin θ_{2}) = 1$
$\bar{z} w$ $= | z | | w | (\cos (- θ_{1}) + i s i n (- θ_{1})) (c o s θ_{2} + i \sin θ_{2})$
$\bar{z} w$ $= \cos [(θ_{2} - θ_{1}) + i s i n (θ_{2} - θ_{1})]$
$= [\cos (- \frac{π}{2}) + i \sin (- \frac{π}{2})]$
$= 1 [0 - i] = - i$

Question:25 i) For any two complex numbers z1, z2 and any real numbers a, b, $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2} = . . . .$ ii) The value of $\sqrt{- 25} * \sqrt{- 9}$ is iii) The number $\frac{(1 - i)^{3}}{1 - i^{3}}$ is equal to ....... iv) The sum of the series $i + i^{2} + i^{3} + . . . + i^{1000}$ upto 1000 terms is ... v) Multiplicative inverse of 1 + i is ................ vi) If $z_{1} a n d z_{2}$ are complex numbers such that $z_{1} + z_{2}$ is a real number, then $z_{2}$ = vii) $a r g (z) + a r g (\bar{z})$ $(\bar{z} \neq 0)$ is .... viii) If $| z + 4 | \leq 3$ then the greatest and least values of |z+1| are ..... and ..... ix) if $\frac{z - 2}{z + 2} = \frac{π}{6}$ then the locus of z is ....... x) If $| z | = 4$ and $a r g (z) = \frac{5 π}{6}$ then z=

Answer:

i) $| a z_{1} - b z_{2} |^{2} + | b z_{1} + a z_{2} |^{2}$
$= | a z_{1} |^{2} + | b z_{2} |^{2} - 2 R e (a z_{1} . b (\bar{z_{2}}) + | b z_{1} |^{2} + | a z_{2} |^{2} + 2 R e (a z_{1} . b \bar{z_{2}}) ?)$
$= | a z_{1} |^{2} + | b z_{2} |^{2} + | b z_{1} |^{2} + | a z_{2} |^{2} = (a^{2} + b^{2}) (| z_{1} |^{2} + | z_{2} |^{2})$
ii) $\sqrt{- 25} * \sqrt{- 9} = 5 i * 3 i = 15 i^{2} = - 15$
iii) $\frac{(1 - i)^{3}}{1 - i^{3}} = \frac{(1 - i)^{3}}{(1 - i) (1 + i + i^{2})}$
$= \frac{(1 - i)^{3}}{(1 + i - 1)} = \frac{1 + i^{2} - 2 i}{} i$
$= \frac{1 - 1 - 2 i}{i} = - \frac{2 i}{i} = - 2$
iv) $i + i^{2} + i^{3} + . . . + i^{1000} = 0 [\sum_{n = 1}^{1000} i^{n} = 0]$
v) $\frac{1}{1 + i} = 1 * \frac{1 - i}{(1 + i) (1 - i)} = \frac{1}{2} (1 - i)$
vi) Let $z_{1} = x_{1} + i y_{1} a n d z_{2} = x_{2} + i y_{2}$
$z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})$
If $z_{1} + z_{2}$ is real then $y_{1} + y_{2} = 0$
$y_{1} = - y_{2}$
$z_{2} = x_{2} - i y_{1}$
$z_{2} = x_{1} - i y_{1} (w h e n x_{1} = x_{2})$
So, $z_{2} = \bar{z_{1}}$
vii) $a r g (z) + a r g (\bar{z})$
$I f a r g (z) = θ, t h e n \arg (\bar{z}) = - θ$
$θ + (- θ) = 0$
viii) $| z + 4 | \leq 3$
$= | z + 4 - 3 | \leq | z + 4 | + | - 3 |$
$= | z + 4 - 3 | \leq 3 + 3$
$= | z + 4 - 3 | \leq 6$
The greatest value is 6 and the least value of $| z + 1 |$ is 0
ix) $\frac{z - 2}{z + 2} = \frac{π}{6}$
Let $z = x + i y$
$| \frac{x + i y - 2}{x + i y + 2} | = | \frac{(x - 2) + i y}{(x + 2) + i y} | = \frac{π}{6}$
$6 | (x - 2) + i y | = π | (x + 2) + i y |$
$6 \sqrt{(x - 2)^{2} + y^{2}} = π \sqrt{(x + 2)^{2} + y^{2}}$
$36 [x^{2} + 4 - 4 x + y^{2}] = π^{2} [x^{2} + 4 + 4 x + y^{2}]$
$36 x^{2} + 144 - 144 x + 36 y^{2} = π^{2} x^{2} + 4 π^{2} + 4 π^{2} x + π^{2} y^{2}$
$(36 - π^{2}) x^{2} + (36 - π^{2}) - (144 + 4 π^{2}) x + 144 - 4 π^{2} = 0$
Which represents an equation of a circle
x) $| z | = 4$ and $a r g (z) = \frac{5 π}{6}$
Let $z = x + i y$
$| z | = \sqrt{x^{2} + y^{2}} = 4$
$x^{2} + y^{2} = 16$
$a r g (z) = \tan^{- 1} \frac{y}{x} = \frac{5 π}{6}$
$\frac{y}{x} = \tan (π - \frac{π}{6})$
$= - \tan \frac{π}{6} = - \frac{1}{\sqrt{3}}$
$x = - \sqrt{3} y$
$(- \sqrt{3} y)^{2} + y^{2} = 16$
$3 y^{2} + y^{2} = 16$
$4 y^{2} = 16$
$y^{2} = 4$
$y = \pm 2$
$x = - 2 \sqrt{3}$
$z = - 2 \sqrt{3} + 2 i$

Question:26 State True or False for the following:
i) The order relation is defined on the set of complex numbers.
ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction
iii) For any complex number z, the minimum value of |z| + |z – 11 is 1
iv) The locus represented by |z — 11= |z — i| is a line perpendicular to the join of the points (1,0) and (0, 1)
v) If z is a complex number such that z ≠ 0 and Re(z) = 0, then Im (z²) = 0
vi) The inequality |z – 4| < |z – 2| represents the region given by x > 3.
(vii) Let Z₁ and Z₂ be two complex numbers such that |z, + z₂| = |z₁ j + |z₂|, then arg (z₁ – z₂) = 0.
(viii) 2 is not a complex number.

Answer:

(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex numbers can be compared. So, it is false.

(ii) Let $z = x + i y$

$z . i = (x + i y) i = x i = - y$ which rotates at angle of 180. So, it is ‘false’.

(iii) Let $z = x + i y$

$| z | + | z - 1 | = \sqrt{x^{2} + y^{2}} + \sqrt{(x - 1)^{2} + y^{2}}$

The value of $| z | + | z - 1 |$ is minimum when $x = 0, y = 0 i . e ., 1$

Hence, it is true.

iv) Let $z = x + i y$

Given that $| z - 1 | = | z - i |$

$| x + i y - 1 | = | x + i y - i |$

$| (x - 1) + i y | = | x - (1 - y) i |$

$\sqrt{(x - 1)^{2} + y^{2}} = {\sqrt{x^{2} + (1 - y)}}^{2}$

$(x - 1)^{2} + y^{2} = x^{2} + (1 - y)^{2}$

$x^{2} - 2 x + 1 + y^{2} = x^{2} + 1 + y^{2} - 2 y$

$- 2 x + 2 y = 0 x - y = 0$ which is a straight line slope=1

Now, the equation of the line through the point 1,0and 0,1

$y - 0 = \frac{1 - 0}{0 - 1} (x - 1)$

$y = - x + 1$ whose slope=-1

Multiplication of the slopes of two lines =-1*1=-1

So, they are perpendicular. Hence, it is true.

v)Let $z = x + i y$ $z \neq 0 a n d R e (z) = 0$

Since, the real part is 0 $x = 0, z = 0 + i y = i y$

$1 m (z^{2}) = y^{2} i^{2} = - y^{2}$ which is real Hence, it is False.

vi) $(z - 4) < | z - 2 |$

Let $z = x + i y$

$| x + i y - 4 | < | x + i y - 2 |$

$| (x - 4) + i y | < | (x - 2) + i y |$

$\sqrt{(x - 4)^{2} + y^{2}} < \sqrt{(x - 2)^{2} + y^{2}}$

$(x - 4)^{2} + y^{2} < (x - 2)^{2} + y^{2}$

$(x - 4)^{2} < (x - 2)$

$x^{2} + 16 - 8 x < x^{2} + 4 - 4 x$

$8 x + 4 x < - 16 + 4$

$- 4 x < - 12$

$x > 3$ Hence, it is true.

vii) Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$

$| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$

$| x_{1} + i y_{1} + x_{2} + i y_{2} | = | (x_{1} + i y_{1}) + (x_{2} + y_{2} i) |$

$= \sqrt{(x_{1} + x_{2})^{2} + (y_{1} + y_{2})^{2}}$

$= \sqrt{x_{1}^{2} + y_{1}^{2}} + \sqrt{x_{2}^{2} + y_{2}^{2}}$

Squaring both sides, we get $(x_{1}^{2} + x_{2}^{2} + 2 x_{1} x_{2} + y_{1}^{2} + y_{2}^{2} + 2 y_{1} y_{2}$ )

$= x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} + 2 \sqrt{x_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + x_{2}^{2} y_{1}^{2} + y_{1}^{2} y_{2}^{2}}$

$= 2 x_{1} x_{2} + 2 y_{1} y_{2} = \sqrt{x_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + x_{2}^{2} y_{1}^{2} + y_{1}^{2} y_{2}^{2}}$

$= x_{1} x_{2} + y_{1} y_{2} = \sqrt{x_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + x_{2}^{2} y_{1}^{2} + y_{1}^{2} y_{2}^{2}}$

Again squaring on both sides we get $x_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + 2 x_{1} y_{1} x_{2} y_{2} = x_{1}^{2} x_{2}^{2} + x_{1}^{2} y_{2}^{2} + x_{2}^{2} y_{1}^{2} + y_{1}^{2} y_{2}^{2} 2 x_{1} y_{1} x_{2} y_{2} = x_{1}^{2} y_{2}^{2} + x_{2}^{2} y_{1}^{2}$

$x_{1}^{2} y_{1}^{2} + x_{2}^{2} y_{1}^{2} - 2 x_{1} y_{1} x_{2} y_{2} = 0$

$(x_{1} y_{2} - x_{2} y_{1})^{2} = 0$

$x_{1} y_{2} - x_{2} y_{1} = 0$

$x_{1} y_{2} = x_{2} y_{1}$

$\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}}$

$\frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}$

$a r g (z_{1}) = a r g (z_{2}) = 0$ Hence, it is false.

(viii) Since every real number is a complex number. So, 2 is a complex number. Hence, it is false.

Question:27

COLUMN A	COLUMN B
a) The polar form of $i + \sqrt{3}$ is	i) Perpendicular bisector of the segment joining (-2,0) and (2,0)
b)The amplitude of $- 1 + \sqrt{- 3}$ is	ii) On or outside the circle having centre at (0,-4) and radius 3.
c) If $\| z + 2 \| = \| z - 2 \|$ then locus of z is	iii) $2 π / 3$
d)If $\| z + 2 i \| = \| z - 2 i \|$ then locus of z is	iv)Perpendicular bisector of the segment joining (0,-2) and (0,2)
e) Region represented by $\| z + 4 i \| \geq 3$ is	v) $2 (\cos \frac{π}{6} + i \sin \frac{π}{6})$
f) Region represented by $\| z + 4 \| \geq 3$ is	vi) On or outside the circle having a centre at (-4,0) and a radius of 3 units.
g)Conjugate of $\frac{1 + 2 i}{1 - i}$ lies in	vii) First Quadrant
h) Reciprocal of 1-i lies in	viii) Third Quadrant

Answer:

a)Given $z = i + \sqrt{3}$

Polar form of z $= r [c o s θ + i s i n θ] = i + \sqrt{3}$ .

$r = \sqrt{3 + 1} = 2$

And $\tan α = \frac{1}{\sqrt{3}}$

$α = \frac{π}{6}$

Since $x > 0, y > 0$

Polar form of z $= 2 [c o s \frac{π}{6} + i s i n \frac{π}{6}]$

b) Given that $z = - 1 + \sqrt{3} = - 1 + \sqrt{3} i$

Here argument z $= \tan^{- 1} | \frac{\sqrt{3}}{- 1} | = \tan^{- 1} \sqrt{3} = \frac{π}{3}$

So, $α = \frac{π}{3}$

Since, x<0 and y>0 $θ = π - α = π - \frac{π}{3} = \frac{2 π}{3}$

c) $| z + 2 | = | z - 2 |$

$| x + i y + 2 | = | x + i y - 2 |$

$| (x - 2) + i y | = | (x - 2) + i y |$

$\sqrt{(x + 2)^{2} + y^{2}} = \sqrt{(x - 2)^{2} + y^{2}}$

$(x + 2)^{2} + y^{2} = (x - 2)^{2} + y^{2}$

$(x + 2)^{2} = (x - 2)^{2}$

$x^{2} + 4 + 4 x = x^{2} + 4 - 4 x$

$8 x = 0, x = 0$

which represents the equation of the y-axis and is perpendicular to the line joining the points (-2,0) and (2,0)

d) $| z + 2 i | = | z - 2 i |$

$(x + i y + 2 i) = | x + i y - 2 i |$

$| x + (y + 2) i | = | x + (y - 2) i |$

$\sqrt{x^{2} + (y + 2)^{2}} = \sqrt{x^{2} + (y - 2)^{2}}$

$x^{2} + (y + 2)^{2} = x^{2} + (y - 2)^{2}$

$(y + 2)^{2} = (y - 2)^{2}$

$y^{2} + 4 + 4 y = y^{2} + 4 - 4 y$

$8 y = 0, y = 0$

which is the equation of the x-axis and is perpendicular to the line segment joining (0,-2)and (0,2)

e) $| z + 4 i | \geq 3$

$| x + i y + 4 | \geq 3$

$| x + (y + 4) i | \geq 3$

$\sqrt{x^{2} + (y + 4)^{2}} \geq 3$

$x^{2} + (y + 4)^{2} \geq 9$

$x^{2} + y^{2} + 8 y + 16 \geq 9$

$x^{2} + y^{2} + 8 y + 7 \geq 0$ $r = \sqrt{(4)^{2} - 7} = 3$

which represents a circle on or outside having a centre (0,-4)

f) $| z + 4 | \leq 3$

Let $z = x + i y$ Then, $| x + i y + 4 | \leq 3$

$| (x + 4) + i y | \leq 3$

$\sqrt{(x + 4)^{2} + y^{2}} \leq 3$

$x^{2} + 8 x + 16 + y^{2} \leq 9$

$x^{2} + y^{2} + 8 x + 7 \leq 0$ which is a circle having centre (-4,0) and $r = \sqrt{(4)^{2} - 7} = \sqrt{9} = 3$ and is on the circle.

g) Let $z = \frac{1 + 2 i}{1 - i}$

$\frac{1 + 2 i}{1 - i} * \frac{1 + i}{1 + i} = \frac{1 + i + 2 i + 2 i^{2}}{1 - i^{2}}$

$= \frac{1 + i + 2 i - 2}{1 + 1} = \frac{- 1 + 3 i}{2}$

$= - \frac{1}{2} + \frac{3}{2} i$ which lies in third quadrant..

h) Given that z=1-i

Reciprocal of $z = \frac{1}{z} = \frac{1}{1 - i} * \frac{1 + i}{1 + i}$

$= \frac{1 + i}{1 + i^{2}}$ which lies in first quadrant.

$S o, (a) - (v), (b) - (i i i), (c) - (i), (d) - (i v), (e) - (i i), (f) - (v i), (g) - (v i i i), (h) - (v i i)$

Question:28 What is the conjugate of $\frac{2 - i}{{(1 - 2 i)}^{2}}$ ?

Answer:

$\frac{2 - i}{{(1 - 2 i)}^{2}}$ $= \frac{2 - i}{1 + 4 i^{2} - 4 i}$
$= \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} * \frac{- 3 + 4 i}{- 3 + 4 i}$
$= \frac{- 6 + 8 i + 3 i - 4 i^{2}}{9 - 16 i^{2}}$
$\frac{- 6 + 11 i + 4}{9 - 16 i^{2}} = \frac{- 2 + 11 i}{9 + 16}$
$\frac{- 2 + 11 i}{25} = \frac{- 2}{25} + \frac{11}{25} i$
Hence, $\bar{z} = - \frac{2}{25} - \frac{11}{25} i$

Question:29 If $| z_{1} | = | z_{2} |$ , is it necessary that $z_{1} = z_{2}$ ?

Answer:

Let $z_{1} = x_{1} + i y_{1} a n d z_{2} = x_{2} + i y_{2}$
$| x_{1} + i y_{1} | = | x_{2} + i y_{2} |$
$= \sqrt{x_{1}^{2} + y_{1}^{2}} = \sqrt{x_{2}^{2} + y_{2}^{2}}$
$x_{1}^{2} + y_{1}^{2} = x_{2}^{2} + y_{2}^{2}$
Consider $z_{1} = 3 + 4 i z_{2} = 4 + 3 i$
$z_{1} \neq z_{2}$ Hence, it is not necessary that $z_{1} = z_{2}$

Question:31 Find z if $| z | = 4$ and $a r g (z) = 5 π / 6$ ..

Answer:

Polar form of $z = r [c o s θ + i \sin θ]$
$= 4 [\cos \frac{5 π}{6} + i \sin \frac{5 π}{6}]$
$= 4 [\cos (π - \frac{π}{6}) + i \sin (π - \frac{π}{6})]$
$= 4 [- \cos (\frac{π}{6}) + i \sin (\frac{π}{6})]$
$= 4 [- \frac{\sqrt{3}}{2} + i \frac{1}{2}]$
$= - 2 \sqrt{3} + 2 i$

Question:32 Find $(1 + i) \frac{(2 + i)}{(3 + i)}$ .

Answer:

$| (1 + i) \frac{(2 + i)}{(3 + i)} * \frac{3 - i}{3 - i} | = | (1 + i) . \frac{6 - 2 i + 3 i - i^{2}}{9 - i^{2}} |$
$= | \frac{(1 + i) (7 + i)}{9 + 1} |$
$= | \frac{7 + i + 7 i + i^{2}}{9 + 1} |$
$= | \frac{7 + i + 7 i + i^{2}}{10} |$
$= | \frac{7 + 8 i - 1}{10} |$
$= | \frac{6 + 8 i}{10} |$
$= | \frac{3}{5} + \frac{4}{5} i |$
$= \sqrt{{(\frac{3}{5})}^{2} + {(\frac{4}{5})}^{2}} = 1$

Question:33 Find the principal argument of $(1 + i \sqrt{3})^{2}$ .

Answer:

$\begin{matrix} (1 + i \sqrt{3})^{2} = 1 + i^{2} \cdot 3 + 2 \sqrt{3} i \\ = 1 - 3 + 2 \sqrt{3} i = - 2 + 2 \sqrt{3} i \\ \tan α = | \frac{2 \sqrt{3}}{- 2} | \\ \tan α = | - \sqrt{3} | = \sqrt{3} \\ \tan α = \tan \frac{π}{3} \\ α = \frac{π}{3} \end{matrix}$

Now Re(z)<0 and image(z)>0 $a r g (z) = π - α = π - \frac{π}{3} = \frac{2 π}{3}$
$Hence, the principal arg = \frac{2 π}{3}$

Question:34 Where does z lie, if $| \frac{z - 5 i}{z + 5 i} | = 1$ .

Answer:

$\begin{matrix} Let z = x + i y | \frac{x + y i - 5 i}{x + i y + 5 i} | = 1 \\ | x + (y - 5) i | = | x + (y + 5) i | \\ x^{2} + (y - 5)^{2} = x^{2} + (y + 5)^{2} \\ (y - 5)^{2} = (y + 5)^{2} \\ \begin{array}{r} y^{2} + 25 - 10 y = y^{2} + 25 + 10 y \\ 20 y = 0 y = 0 \end{array} \end{matrix}$

Hence, z lies on the x-axis i.e., the real axis.

Question:35 $\sin x + i \cos 2 x$ and $\cos x - i \sin 2 x$ are conjugate to each other for: A. $x = n π$ B. $x = (n + \frac{1}{2}) \frac{π}{2}$ C. $x = 0$ D. No value of x

Answer:

The answer is option (c).
let $z = \sin x + i \cos 2 x \bar{z} = \sin x - i \cos 2 x$

But we are given that $\bar{z} = \cos x - i \sin 2 x$

$\sin x - i \cos 2 x = \cos x - i \sin 2 x$

Comparing the real and imaginary parts, we get

$\sin x = \cos x$ and $\cos 2 x = \sin 2 x$

$\tan x = 1 and \tan 2 x = 1$
$\tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x} = 1$
The above value is not satisfied by tan x = 1.
Hence, no value of x is possible.

Question:36 The real value of α for which the expression $\frac{(1 - i \sin α)}{1 + 2 i \sin α}$ is purely real is: A. $(n + 1) \frac{π}{2}$ B. $(2 n + 1) \frac{π}{2}$ C. $n π$ D. None of these

Answer:

The answer is the option (c).
Let $z = \frac{(1 - i \sin α)}{1 + 2 i \sin α}$
$\begin{aligned} \begin{array}{c} = \frac{(1 - i \sin α) (1 - 2 i \sin α)}{(1 + 2 i \sin α) (1 - 2 i \sin α)} \\ = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{(1)^{2} - (2 i \sin α)^{2}} \\ = \frac{1 - 3 i \sin α - 2 \sin^{2} α}{1 - 4 i^{2} \sin^{2} α} = \frac{(1 - 2 \sin^{2} α) - 3 i \sin α}{1 + 4 \sin^{2} α} \\ \frac{- 3 i \sin α}{1 + 4 \sin^{2} α} = 0 \\ \sin α = 0 α = n π \end{array} \\ Hence, c is correct. \end{aligned}$

Question:37 If z = x + iy lies in the third quadrant, the $\frac{\bar{z}}{z}$ also lies in the third quadrant if A. x > y > 0 B. x < y < 0 C. y < x < 0 D. y > x > 0

Answer:

The answer is the option (b).
If z lies in the third quadrant,So, $x < 0 a n d y < 0, z = x + i y$
$\begin{aligned} \frac{\bar{z}}{z} & = \frac{x - i y}{x + i y} = \frac{x - i y}{x + i y} * \frac{x - i y}{x - i y} \\ = \frac{x^{2} + i^{2} y^{2} - 2 x y i}{x^{2} - i^{2} y^{2}} \\ = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} - \frac{2 x y}{x^{2} + y^{2}} i \end{aligned}$
When z lies in the third quadrant then $\frac{\bar{z}}{z}$ will also be in the third quadrant.
$\begin{matrix} \frac{x^{2} - y^{2}}{x^{2} + y^{2}} < 0 and \frac{2 x y}{x^{2} + y^{2}} > 0 \\ x^{2} - y^{2} < 0 and 2 x y > 0 \\ x^{2} < y^{2} and x y > 0 \\ So, x < y < 0 \end{matrix}$
Hence, b is correct.

Question:38 The value of $(z + 3) (\bar{z} + 3)$ is equivalent to A. $| z + 3 |^{2}$ B. |z – 3| C. $z^{2} + 3$ D. None of these

Answer:

The answer is the option (a).
$L e t z = x + i y S o, (z + 3) (\bar{z} + 3) = (x + i y + 3) (x - i y + 3)$
$= [(x + 3) + i y] [(x + 3) - i y]$
$= (x + 3)^{2} - y^{2} i^{2} = (x + 3)^{2} + y^{2}$
$= | x + 3 + i y |^{2} = | z + 3 |^{2}$

Question:39 If ${(\frac{1 + i}{1 - i})}^{x} = 1$ , then A. $x = 2 n + 1$ B. $x = 4 n$ C. $x = 2 n$ D. $x = 4 n + 1$

Answer:

The answer is the option(b).
$\begin{matrix} {(\frac{1 + i}{1 - i})}^{x} = 1 \\ {(\frac{(1 + i) (1 + i)}{(1 - i) (1 + i)})}^{x} = 1 \\ {(\frac{1 - 1 + 2 i}{1 + 1})}^{x} = 1 \\ {(\frac{2 i}{2})}^{x} = 1 (i)^{x} = 1 \end{matrix}$
$x = 4 n, n ϵ N$
Hence, the correct option is

Question 40: A real value of x satisfies the equation $\frac{3 - 4 i x}{3 + 4 i x} = α - i β$ ( $α, β \in R$ ) , if $α^{2} + β^{2} =$ A. 1 B. –1 C. 2 D. –2

Answer:

The answer is the option (a).

$\begin{aligned} Given that \frac{3 - 4 i x}{3 + 4 i x} = α - i β \\ = \frac{3 - 4 i x}{3 + 4 i x} * \frac{3 - 4 i x}{3 - 4 i x} = α - i β \\ \frac{9 - 12 i x - 12 i x + 16 i^{2} x^{2}}{9 - 16 i^{2} x^{2}} = α - i β \\ = \frac{9 - 16 x^{2}}{9 + 16 x^{2}} - \frac{24 x}{9 + 16 x^{2}} = α - i β \dots \dots . (i) \\ \frac{9 - 16 x^{2}}{9 + 16 x^{2}} + \frac{24 x}{9 + 16 x^{2}} = α + i β \dots \dots \dots (i i) \end{aligned}$
Multiplying eqn (i )and (ii )we get
${(\frac{9 - 16 x^{2}}{9 + 16 x^{2}})}^{2} + {(\frac{24 x}{9 + 16 x^{2}})}^{2} = α^{2} + β^{2}$
$\frac{81 + 256 x^{4} - 288 x^{2} + 576 x^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2}$
$\begin{array}{r} \frac{81 + 256 x^{4} + 288 x^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} \\ \frac{{(9 + 16 x^{2})}^{2}}{{(9 + 16 x^{2})}^{2}} = α^{2} + β^{2} = 1 \end{array}$

Question:41 Which of the following is correct for any two complex numbers $z_{1} a n d z_{2}$ ? A. $| z_{1} z_{2} | = | z_{1} | | z_{2} |$ B. $a r g (z_{1} z_{2}) = a r g (z_{1}) . A r g (z_{2})$ C. $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ D. $| z_{1} + z_{2} | \geq | z_{1} | - | z_{2} |$

Answer:

The answer is the option (a).
$\begin{matrix} z_{1} = r_{1} (\cos θ_{1} + i \sin θ_{2}) \\ | z_{1} | = r_{1} \\ z_{2} = r_{2} (\cos θ_{2} + i \sin θ_{2}) \\ | z_{2} | = r_{2} \\ z_{1} z_{2} = r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] \\ = r_{1} (\cos θ_{1} + i \sin θ_{1}) \cdot r_{2} (\cos θ_{2} + i \sin θ_{2}) \end{matrix}$
$\begin{matrix} = r_{1} r_{2} [(\cos θ_{1} \cos θ_{2} - \sin θ_{1} \sin θ_{2}) + i (\sin θ_{1} \cos θ_{2} + \cos θ_{1} \sin θ_{2})] \\ = r_{1} r_{2} [\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] | z_{1} z_{2} | = | z_{1} ‖ z_{2} | \end{matrix}$

Hence, option A is correct.

Question:42 The point represented by the complex number $2 - i$ is rotated about the origin through an angle $π / 2$ in the clockwise direction, the new position of a point is: A. $1 + 2 i$ B. $- 1 - 2 i$ C. $2 + i$ D. $- 1 + 2 i$

Answer:

The answer is the option (b).
If z rotated through an angle of $π / 2$ about the origin in a clockwise direction.
Then the new position $= z . e^{- (\frac{π}{2})}$
$= (2 - i) . e^{- (\frac{π}{2})}$
$= (2 - i) . [\cos (- \frac{π}{2}) + i s i n (- \frac{π}{2})]$
$= (2 - i) . (0 - i)$
$= - 1 - 2 i$
Hence, the correct option is (b)

Question:43 Let $x, y \in R$ , then $x + i y$ is a non-real complex number if: A. $x = 0$ B. $y = 0$ C. $x \neq 0$ D. $y \neq 0$

Answer:

The answer is the option (d).
$x + i y$ is a non-real complex number if $y \neq 0$ ,

Hence, d is correct

Question:44 If $a + i b = c + i d$ , then A. $a^{2} + c^{2} = 0$ B. $b^{2} + c^{2} = 0$ C. $b^{2} + d^{2} = 0$ D. $a^{2} + b^{2} = c^{2} + d^{2}$

Answer:

The answer is the option (d).
$a + i b = c + i d$
$| a + i b | = | c + i d |$
$\sqrt{a^{2} + b^{2}} = \sqrt{c^{2} + d^{2}}$
Squaring both sides, $a^{2} + b^{2} = c^{2} + d^{2}$
Hence, d is correct

Question:45 The complex number z which satisfies the condition $| \frac{i + z}{i - z} | = 1$ lies on A. circle $x^{2} + y^{2} = 1$ B. the x-axis C. the y-axis D. the line $x + y = 1$

Answer:

The answer is the option (b).
$\begin{matrix} Let z = x + i y | \frac{i + x + i y}{i - x - y i} | = 1 \\ | \frac{x - (y + 1) i}{- x - (y - 1) i} | = 1 \\ | x + (y + 1) i | = | - x - (y - 1) i | \\ \sqrt{x^{2} + (y + 1)^{2}} = \sqrt{x^{2} + (y - 1)^{2}} \\ x^{2} + (y + 1)^{2} = x^{2} + (y - 1)^{2} \\ y^{2} + 2 y + 1 = y^{2} - 2 y + 1 \\ 2 y = - 2 y \\ y = 0; x - a x i s \end{matrix}$

Hence, b is correct

Question:46 If z is a complex number, then A. $| z^{2} | > | z |^{2}$ B. $| z^{2} | e q u a l s | z |^{2}$ C. $| z^{2} | < | z |^{2}$ D. $| z^{2} | \geq | z |^{2}$

Answer:

. The answer is the option (b).
$L e t z = x + i y, | z |^{2} = x^{2} + y^{2}$
$N o w, z^{2} = x^{2} + y^{2} i^{2} + 2 x y i$
$z^{2} = x^{2} - y^{2} + 2 x y i$
$| z |^{2} = \sqrt{(x^{2} - y^{2})^{2} + (2 x y)^{2}}$
$= \sqrt{x^{4} + y^{4} + 2 x^{2} y^{2}} = \sqrt{(x^{2} + y^{2})^{2}}$
$| z |^{2} = x^{2} + y^{2} = | z |^{2}$

Hence, b is correct

Question:47 $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$ is possible if A. $z_{1} = \bar{z_{1}}$ B. z2=z1 C. $a r g (z_{1}) = a r g (z_{2})$ D. $| z_{1} | = | z_{2} |$

Answer:

The answer is the option (c).
Since, $| z_{1} + z_{2} | = | z_{1} | + | z_{2} |$
$| z_{1} + z_{2} | = | r_{1} (\cos θ_{1} + i \sin θ_{1}) + r_{2} (\cos θ_{2} + i \sin θ_{2}) |$
$= \sqrt{r_{1}^{2} \cos^{2} θ_{1} + r_{2}^{2} \cos^{2} θ_{2} + 2 r_{1} r_{2} \cos θ_{1} \cos θ_{2} + r_{1}^{2} \sin^{2} θ_{1} + r_{2}^{2} \sin^{2} θ_{2} + 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2}}$
So, $\begin{matrix} = \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2})} \\ | z_{1} + z_{2} | = | z_{1} | + | z_{2} | \\ \sqrt{r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2})} = r_{1} + r_{2} \end{matrix}$
Squaring both sides,
$\begin{matrix} r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = r_{1}^{2} + r_{2}^{2} + 2 r_{1} r_{2} \\ 2 r_{1} r_{2} - 2 r_{1} r_{2} \cos (θ_{1} - θ_{2}) = 0 \\ 1 - \cos (θ_{1} - θ_{2}) = 0 \\ \cos (θ_{1} - θ_{2}) = 1 \\ θ_{1} - θ_{2} = 0 \\ θ_{1} = θ_{2} \\ so \arg (z_{1}) = \arg (z_{2}) \end{matrix}$

Question:48 The real value of θ for which the expression $\frac{1 + i \cos θ}{1 - 2 i \cos θ}$ is a real number is: A. $n π + \frac{π}{4}$ B. $n π + (- 1)^{n} \frac{π}{4}$ C. $2 n π + \frac{π}{2}$ D. none of these

Answer:

Let $z = \frac{1 + i \cos θ}{1 - 2 i \cos θ} = \frac{1 + i \cos θ}{1 - 2 i \cos θ} * \frac{1 + 2 i \cos θ}{1 + 2 i \cos θ}$

On solving, we get
$\frac{1 + 3 i \cos θ - 2 \cos^{2} θ}{1 - 4 i^{2} \cos^{2} θ} = \frac{1 + 3 i \cos θ - 2 \cos^{2} θ}{1 + 4 \cos^{2} θ}$
$= \frac{(1 - 2 \cos^{2} θ) + 3 i \cos θ}{1 + 4 \cos^{2} θ}$
If z is a real number, then
$\frac{3 \cos θ}{1 + 4 \cos^{2} θ} = 0$
$3 \cos θ = 0 \cos θ = 0$
$θ = \frac{(2 n + 1) π}{2}, n \in N$

Question:49 The value of $a r g (x)$ when $x < 0$ is: A. 0 B. $\frac{π}{2}$ C. π D. none of these

Answer:

The answer is option (c).
$L e t z = - x + 0 i a n d x < 0 | z | = \sqrt{(- 1)^{2} + (0)^{2}} = 1, x < 0$

Since, the point (-x,0)lies on the negative side of the real axis, principal argument (z)=π

Hence, option c is correct.

Question:50 If $f (z) = \frac{7 - z}{1 - z^{2}}$ where $z = 1 + 2 i$ , then $| f (z) |$ is A. $\frac{| z |}{2}$ B. $| z |$ C. $2 | z |$ D. none of these

Answer:

The answer is option (a).
Given that $z = 1 + 2 i | z | = \sqrt{(1)^{2} + (2)^{2}} = \sqrt{5}$
Now $f (z) = \frac{7 - z}{1 - z^{2}}$
$\begin{aligned} = \frac{7 - (1 + 2 i)}{1 - (1 + 2 i)^{2}} & = \frac{7 - 1 - 2 i}{1 - 1 - 4 i^{2} - 4 i} \\ = \frac{6 - 2 i}{4 - 4 i} & = \frac{3 - i}{2 - 2 i} \end{aligned}$
$\begin{aligned} = \frac{3 - i}{2 - 2 i} * \frac{2 + 2 i}{2 + 2 i} = \frac{6 + 6 i - 2 i - 2 i^{2}}{4 - 4 i} \\ = \frac{6 + 4 i + 2}{4 + 4} & = \frac{8 + 4 i}{8} \\ = 1 + \frac{1}{2} i \end{aligned}$
$f (z) = \sqrt{(1)^{2} + {(\frac{1}{2})}^{2}} = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} = \frac{| z |}{2}$

Hence, option A is correct.

NCERT Solutions for Class 11 Mathematics Chapters

NCERT Solutions for Class 11 Maths: Chapter Wise

NCERT Solutions for Class 11 Mathematics Chapter 1: Sets

NCERT Solutions for Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Solutions for Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Solutions for Class 11 Mathematics Chapter 4: Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Mathematics Chapter 5: Linear Inequalities

NCERT Solutions for Class 11 Mathematics Chapter 6: Permutations and Combinations

NCERT Solutions for Class 11 Mathematics Chapter 7: Binomial Theorem

NCERT Solutions for Class 11 Mathematics Chapter 8: Sequences and Series

NCERT Solutions for Class 11 Mathematics Chapter 9: Straight Lines

NCERT Solutions for Class 11 Mathematics Chapter 10: Conic Sections

NCERT Solutions for Class 11 Mathematics Chapter 11: Introduction to Three-Dimensional Geometry

NCERT Solutions for Class 11 Mathematics Chapter 12: Limits and Derivatives

NCERT Solutions for Class 11 Mathematics Chapter 13: Statistics

NCERT Solutions for Class 11 Mathematics Chapter 14: Probability

Important Notes from NCERT Exemplar Class 11 Maths Solution Chapter 5

Various laws need to be followed and derived by a mathematician when it comes to multiplying two complex numbers. All of them are named after the features of the specific formula derivation. Ahead, you will also learn about the Modulus and Conjugate of complex numbers and how to solve sums with their help. There is an Argand and Polar representation in the chapter, relating to graphical presentations. A point needs to be derived at the ‘x’ and ‘y’ axis to form an Argand or complex plane.

Students, with the help of NCERT Exemplar Class 11 Maths Solutions Chapter 5, will not face any issues when trying to solve such problems. It will also help them score better in exams.

All the concepts have been covered in the NCERT Exemplar Solutions for Class 11 Maths chapter 5. By using the NCERT Exemplar Class 11 Maths chapter 5 solutions PDF Download function, students can access quality study material that is effectively constructed by experts for the best learning experience.

NCERT Solutions for Class 11 Mathematics Chapters

NCERT Exemplar Class 11 Mathematics Chapter 1: Sets

NCERT Exemplar Class 11 Mathematics Chapter 2: Relations and Functions

NCERT Exemplar Class 11 Mathematics Chapter 3: Trigonometric Functions

NCERT Exemplar Class 11 Mathematics Chapter 4: Principles of Mathematical Induction

NCERT Exemplar Class 11 Mathematics Chapter 5: Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Mathematics Chapter 6: Linear Inequalities

NCERT Exemplar Class 11 Mathematics Chapter 7: Permutations and Combinations

NCERT Exemplar Class 11 Mathematics Chapter 8: Binomial Theorem

NCERT Exemplar Class 11 Mathematics Chapter 9: Sequence and Series

NCERT Exemplar for Class 11 Mathematics Chapter 10: Straight Lines

NCERT Exemplar Class 11 Mathematics Chapter 11: Conic Sections

NCERT Exemplar Class 11 Mathematics Chapter 12: Introduction to three-Dimensional Geometry

NCERT Exemplar Class 11 Mathematics Chapter 13: Limits and Derivatives

NCERT Exemplar Class 11 Mathematics Chapter 14: Mathematical Reasoning

NCERT Exemplar Class 11 Mathematics Chapter 15: Statistics

NCERT Exemplar Class 11 Mathematics Chapter 16: Probability

Importance of solving NCERT questions for class 11, Chapter 5, complex numbers and quadratic equations:

Complex numbers and quadratic equations are not only useful chapters in class 11 but also for higher studies and competitive exams. Strengthening basic concepts is a necessity for students so that later they do not face any difficulties solving Complex numbers and quadratic equations questions in higher studies or competitive examinations.
Some important facts about solving Complex numbers and quadratic equations in class 11 are listed below.

Students can study strategically at their own pace after accessing Class 11 Maths NCERT Solutions chapter 5. This will boost their confidence to attempt other questions from this chapter.
Class 11 Maths Chapter 5 NCERT solutions are solved by subject-matter experts and are very reliable at the same time. The solutions provide shortcuts as well as detailed explanations with necessary formulae that will help students to understand the answers better.
These solutions will help students manage their time efficiently in this chapter and understand which questions are easier to approach and which are time-consuming. This will be helpful during the exam.
NCERT solutions for class 11 Maths chapter 5 Complex numbers and quadratic equations is designed to give the students step-by-step solutions for a particular question.

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Frequently Asked Questions (FAQs)

1. What is the significance of Argand Plane in Class 11 Maths Chapter 5?

The Argand Plane is significant in Class 11 Maths Chapter 5 (Complex Numbers) as it provides a geometric representation of complex numbers. Each complex number z=x+iy is plotted as a point (x,y) in a 2D plane, where the horizontal axis represents the real part and the vertical axis represents the imaginary part. This visual approach helps understand complex number operations like addition, subtraction, modulus, and conjugation more intuitively.

2. What are the conjugate and reciprocal of a complex number?

The conjugate of a complex number z=x+i y is $\bar{z}=x-i y$ . It reflects the number across the real axis on the Argand Plane and is useful in division and simplification.

The reciprocal of $z$ is given by:
$\frac{1}{z}=\frac{\bar{z}}{|z|^2}=\frac{x-i y}{x^2+y^2}$
This expression helps divide complex numbers by rationalizing the denominator using the conjugate.

3. What are the important topics covered in NCERT Exemplar Class 11 Maths Chapter 5?

Important topics covered in NCERT Exemplar Class 11 Maths Chapter 5 (Complex Numbers and Quadratic Equations) include:

Algebra of complex numbers (addition, subtraction, multiplication, division)
Modulus and argument of complex numbers
Polar and exponential form
Geometrical representation on Argand Plane
Solving quadratic equations with complex roots
Use of identities and inequalities involving complex numbers

4. How to find the modulus and argument of a complex number?

To find the modulus and argument of a complex number z=x+i y :
- Modulus is the distance from the origin:

$|z|=\sqrt{x^2+y^2}$

- Argument is the angle $θ$ made with the positive real axis:

$\arg (z)=\tan ^{-1}\left(\frac{y}{x}\right)$
The modulus shows the length, and the argument shows the direction of the complex number on the Argand Plane.

5. What is the polar form of a complex number in NCERT Exemplar Class 11 Maths?

In NCERT Exemplar Class 11 Maths, the polar form of a complex number expresses z=x+i y in terms of its modulus $r=\sqrt{x^2+y^2}$ and argument $\theta=\tan ^{-1}\left(\frac{y}{x}\right)$ . It is written as:
$z=r(\cos \theta+i \sin \theta)$
This form is useful for multiplication, division, and finding powers/roots of complex numbers. It provides a clear geometric interpretation using the Argand Plane.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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NCERT Exemplar Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Mathematics Chapters

Important Notes from NCERT Exemplar Class 11 Maths Solution Chapter 5

NCERT Solutions for Class 11 Mathematics Chapters

Importance of solving NCERT questions for class 11, Chapter 5, complex numbers and quadratic equations:

NCERT Exemplar Class 11 Solutions":

Frequently Asked Questions (FAQs)

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